June 2015 MS - C2 Edexcel PDF

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Mark Scheme (Results) Summer 2015

Pearson Edexcel GCE in Core Mathematics C2 (6664/01)

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Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at www.edexcel.com or www.btec.co.uk. Alternatively, you can get in touch with us using the details on our contact us page at www.edexcel.com/contactus.

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Summer 2015 Publications Code UA041196 All the material in this publication is copyright © Pearson Education Ltd 2015

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General Marking Guidance

 All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last.  Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions.  Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie.  There is no ceiling on achievement. All marks on the mark scheme should be used appropriately.  All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate’s response is not worthy of credit according to the mark scheme.  Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited.  Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

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PEARSON EDEXCEL GCE MATHEMATICS General Instructions for Marking 1. The total number of marks for the paper is 75 2. The Edexcel Mathematics mark schemes use the following types of marks: 

M marks: Method marks are awarded for ‘knowing a method and attempting to apply it’, unless otherwise indicated.



A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned.



B marks are unconditional accuracy marks (independent of M marks)



Marks should not be subdivided.

3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. 

bod – benefit of doubt



ft – follow through



the symbol



cao – correct answer only



cso - correct solution only. There must be no errors in this part of the question to obtain this mark



isw – ignore subsequent working



awrt – answers which round to



SC: special case



oe – or equivalent (and appropriate)



d… or dep – dependent



indep – independent



dp decimal places



sf significant figures



 The answer is printed on the paper or ag- answer given



will be used for correct ft

or d… The second mark is dependent on gaining the first mark

4. All A marks are ‘correct answer only’ (cao.), unless shown, for example, as A1 ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks.

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5. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. 6. If a candidate makes more than one attempt at any question:  If all but one attempt is crossed out, mark the attempt which is NOT crossed out.  If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt. 7. Ignore wrong working or incorrect statements following a correct answer.

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General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles).

Method mark for solving 3 term quadratic: 1. Factorisation

( x2  bx  c)  ( x  p)( x  q), where pq  c , leading to x = … ( ax2  bx  c)  ( mx  p)( nx  q), where pq  c and mn  a , leading to x = … 2. Formula Attempt to use the correct formula (with values for a, b and c).

3. Completing the square 2

Solving x  bx  c  0 : 2

b   x    q  c  0, q  0 , leading to x = … 2 

Method marks for differentiation and integration:

1. Differentiation Power of at least one term decreased by 1. ( x  x n

n1

)

2. Integration Power of at least one term increased by 1. ( x  x n

n1

)

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Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent examiners’ reports is that the formula should be quoted first. Normal marking procedure is as follows: Method mark for quoting a correct formula and attempting to use it, even if there are small errors in the substitution of values. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may be lost if there is any mistake in the working. Exact answers Examiners’ reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals.

7

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May 2015 6664 Core Mathematics C2 Mark Scheme Question Number

Scheme

Marks

10

x  2   4 

1.

 10  1   10  1  2 10   2 9   x   2 8   x  1  4   2  4 

Way 1

2

For either the x term or the x 2 term including a correct binomial coefficient with a correct power of x

 ...

First term of 1024

 1024 1280 x  720 x

Way 2

2

Either  1280x or 720x2 (Allow +-1280 x here) Both  1280x and 720x2 (Do not allow +-1280x here)

M1 B1 A1 A1 [4]

 x 10  9  x  2  x  k  2    2 1  10       4 8 2 8     1024(1 .......)

M1

 1024 1280 x  720x 2

B1A1 A1 [4]

10

Notes M1: For either the x term or the x term having correct structure i.e. a correct binomial coefficient in any form with the 2

correct power of x. Condone sign errors and condone missing brackets and allow alternative forms for binomial

 10  10  C1 or   or even   or 10. The powers of 2 or of ¼ may be wrong or missing.  1 1 B1: Award this for 1024 when first seen as a distinct constant term (not 1024x0) and not 1 + 1024 A1: For one correct term in x with coefficient simplified. Either -1280x or 720x 2 ( allow +-1280x here) coefficients e.g.

10

2

 x  with no negative sign. So use of + sign throughout could give M1 B1 A1 A0  4

Allow 720x2 to come from 

A1: For both correct simplified terms i.e. -1280x and 720x2 (Do not allow +-1280x here) 2 Allow terms to be listed for full marks e.g. 1024, 1280 x,  720x N.B. If they follow a correct answer by a factor such as 512 640 x  360x2 then isw Terms may be listed. Ignore any extra terms. Notes for Way 2 M1: Correct structure for at least one of the underlined terms. i.e. a correct binomial coefficient in any form with the correct power of x. Condone sign errors and condone missing brackets and allow alternative forms for binomial coefficients e.g.

10

 10  10  or even   or 10. k may even be 0 or 2k may not be seen. Just consider the bracket for   1 1

C1 or 

this mark. B1: Needs 1024(1…. To become 1024 A1, A1: as before

8

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Question Number 2 (a)

Scheme Way 1 ( x m2)  ( y  1)  k , k > 0 2

Way 2

2

x  y m4 x  2 y  c  0

M1

4  (5)  4  4  2 5  c  0

M1 A1

2

Attempts to use r  (4  2)  (5  1) 2

Marks

2

2

Obtains ( x  2) 2  ( y 1) 2  20

2

2

2

x2  y2  4x  2 y  15  0

(3)

N.B. Special case: ( x  2) ( y 1)  20 is not a circle equation but earns M0M1A0 2

(b) Way 1

2

Gradient of radius from centre to (4, -5) = -2 Tangent gradient = 

(must be correct)

B1

1 their numerical gradient of radius

M1

Equation of tangent is ( y  5)  ' 12 '( x  4) So equation is x – 2y –14 = 0 (or 2y – x +14 = 0 or other integer multiples of this answer) b)Way 2

Quotes xx  yy   2( x  x )  ( y  y) 15  0 and substitutes (4, -5)_ 4 x 5 y  2( x  4)  ( y  5) 15  0 so 2x – 4y – 28 = 0 (or alternatives as in Way 1)

b)Way 3

Use differentiation to find expression for gradient of circle Either 2( x  2)  2( y  1)

dy dx

 0 or states y  1  20  ( x  2) so 2

dy (x  2)  dx 20  ( x  2) 2

Substitute x = 4, y = -5 after valid differentiation to give gradient =

M1 A1 (4) B1 M1,M1A1 (4) B1 M1 M1 A1

Then as Way 1 above ( y  5)  ' 12 '( x  4) so x – 2y –14 = 0

(4) [7]

Notes (a) M1: Uses centre to write down equation of circle in one of these forms. There may be sign slips as shown. 2

M1: Attempts distance between two points to establish r (independent of first M1)- allow one sign slip only using distance formula with -5 or -1, usually (– 5 – 1) in 2nd bracket. Must not identify this distance as diameter. This mark may alternatively (e.g. way 2)be given for substituting (4, -5) into a correct circle equation with one unknown Can be awarded for r 

20 or for r 2  20 stated or implied but not for r 2  20 or r = 20 or r  5 2

A1: Either of the answers printed or correct equivalent e.g. (x  2)2  ( y  1)2  (2 5) 2 is A1 but 2 5 (no bracket) is A0

unless there is recovery Also (x  2)2  ( y  (1))2  (2 5) 2 may be awarded M1M1A1as a correct equivalent. N.B. ( x  2) 2  ( y 1) 2  40commonly arises from one sign error evaluating r and earns M1M1A0 (b) Way 1: B1: Must be correct answer -2 if evaluated (otherwise may be implied by the following work) M1: Uses negative reciprocal of their gradient M1: Uses y  y1  m( x  x1 ) with (4,-5) and their changed gradient or uses y = mx + c and (4, -5) with their changed gradient (not gradient of radius) to find c A1: answers in scheme or multiples of these answers (must have “ = 0” ). NB Allow 1x – 2y –14 = 0 N.B. ( y  5)  ' 12 '( x  4) following gradient of is ½ after errors leads to x – 2y –14 = 0 but is worth B0M0M0A0 Way 2: Alternative method (b) is rare. Way 3: Some may use implicit differentiation to differentiate- others may attempt to make y the subject and use chain rule B1: the differentiation must be accurate and the algebra accurate too. Need to take (-) root not (+)root in the alternative M1: Substitutes into their gradient function but must follow valid accurate differentiation M1: Must use “their” tangent gradient and y  5  m( x  4) but allow over simplified attempts at differentiation for this mark. A1: As in Way 1

9

PhysicsAndMathsTutor.com Question Number

Scheme

Marks

3. Way 1 (a)

f ( x)  6 x 3  3 x 2  Ax  B Attempting f (1) = 45 or f ( 1)  45 f ( 1)   6  3  A  B  45 or  3  A  B  45  B  A  48 * (allow 48 = B – A)

Way 1 (b)

Attempting f (  12 )  0

M1 A1 * cso (2) M1

6  12   3  12   A  12   B  0 or  12 A  B  0 or A = 2B 3

2

A1 o.e. M1 A1

Solve to obtain B =  48 and A =  96 Way 2 (a)

Way 2 (b)

(4)

Long Division (6 x3  3 x2  Ax  B) ( x 1)  6 x2  px  q and sets remainder = 45

M1

Quotient is 6 x2  3x  ( A  3) and remainder is B – A – 3 =45 so B – A = 48 *

A1*

(6 x  3 x  Ax  B) (2 x 1)  3x  px  q and sets remainder = 0

M1

A A and remainder is B  = 0 2 2

A1

3

2

2

2 Quotient is 3x 

Then Solve to obtain B =  48 and A =  96 as in scheme above (Way 1) (c)

Obtain

 3x

2

 48

 , x

2

 16

 , 6x

2

 96

 ,  3x

2





A 2

A

M1 A1 B

 2  2 2  , 3x  B  , x   or  x   as B1ft 6 3    

factor or as quotient after division by (2x + 1) . Division by (x+4) or (x-4) see below









 



Factorises 3 x2  48 , x2 16 , 48  3 x2 , 16  x2 or 6x2  96

M1

= 3 (2x + 1)(x + 4)(x  4) (if this answer follows from a wrong A or B then award A0) A1cso isw if they go on to solve to give x = 4, -4 and -1/2 (3) [9] Notes (a) Way 1: M1: 1 or –1 substituted into f(x) and expression put equal to ±45 A1*: Answer is given. Must have substituted  1 and put expression equal to +45. Correct equation with powers of  1 evaluated and conclusion with no errors seen. Way 2: M1: Long division as far as a remainder which is set equal to ±45 A1*: See correct quotient and correct remainder and printed answer obtained with no errors (b) Way 1: M1: Must see f ( 12 ) and “= 0” unless subsequent work implies this. A1: Give credit for a correct equation even unsimplified when first seen, then isw. A correct equation implies M1A1. M1: Attempts to solve the given equation from part (a) and their simplified or unsimplified linear equation in A and B from part (b) as far as A =... or B = ...(must eliminate one of the constants but algebra need not be correct for this mark). May just write down the correct answers. A1: Both A and B correct Way 2: M1: Long division as far as a remainder which is set equal to 0 A1: See correct quotient and correct remainder put equal to 0 M1A1: As in Way 1 There may be a mixture of Way 1 for (a) and Way 2 for (b) or vice versa. (c) B1: May be written straight down or from long division, inspection, comparing coefficients or pairing terms M1: Valid attempt to factorise a listed quadratic (see general notes) so  3x  16 x  3 could get M1A0

A1cso: (Cannot be awarded if A or B is wrong) Needs the answer in the scheme or -3 (2x+1)(4+x)(4  x) or equivalent but factor 3 must be shown and there must be all the terms together with brackets.



 



Way 2: A minority might divide by (x- 4) or (x + 4) obtaining 6 x2  27 x  12 or 6 x2  21x 12



 

2 2 They then need to factorise 6 x  27 x  12 or 6 x  21x  12



for B1

for M1

Then A1cso as before Special cases: If they write down f(x) = 3 (2x+1)(x+4)(x - 4) with no working, this is B1 M1 A1 But if they give f(x) = (2x+1)(x+4)(x - 4) with no working (from calculator?) give B1M0A0 And f(x) = (2x + 1)(3x + 12)(x  4) or f(x) = (6x + 3)(x + 4)(x  4) or f(x) = (2x + 1)(x + 4)(3x  12) is B1M1A0 10

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Question Number 4.(a)

Scheme In triangle OCD complete method used to find angle COD so:

µ  Either cos COD

82  82  72 or uses COD  2  arcsin 3.5 8 oe 2 8 8

( COD  0.9056(331894) ) (b)

= 0.906 (3sf) *

Uses s = 8  for any  in radians or

 (c)

Marks

  "COD " 2

 awrt 1.12  or

 360

so COD 

M1

accept awrt 0.906

A1 *

 2  8 for any  in degrees

M1

2  awrt 2.24 and Perimeter = 23+( 16  )

accept awrt 40.9 (cm) Either Way 1: (Use of Area of two sectors + area of triangle) Area of triangle = 12  8  8  sin 0.906 (or 25.1781155 accept awrt 25.2)or 1 2

2

M1 A1 (3) M1

 8  7  sin1.118 or 12  7  h after h calculated from correct Pythagoras or trig.

Area of sector = 12 8 "1.117979732" (or 35.77535142

(2)

accept awrt 35.8 ) 2

Total Area = Area of two sectors + area of triangle =awrt 96.7 or 96.8 or 96.9 ( cm )

M1 A1 (3)

Or Way 2: (Use of area of semicircle – area of segment) Area of semi-circle = 12    8  8 ( or 100.5) Area of segment =

1 2

M1

8 2  ("0.906" sin"0.906") ( or 3.807)

M1 A1

So area required = awrt 96.7 or 96.8 or 96.9 ( cm2 )

(3) [8]

Notes (a) M1: Either use correctly quoted cosine rule – may quote as 7  8  8  2 8 8 cos    ..... Or split isosceles triangle into two right angled triangles and use arcsin or longer methods using Pythagoras ). There are many ways of showing this result. and arcos (i.e.   2  arccos 3.5 8 2

2

2

Must conclude that COD  A1*: (NB this is a given answer) If any errors or over-approximation is seen this is A0. It needs correct work leading to stated answer of 0.906 or awrt 0.906 for A1. The cosine of COD is equal to 79/128 or awrt 0.617. Use of 0.62 (2sf) does not lead to printed answer. They may give 51.9 in degrees then convert to radians. This is fine. The minimal solution 7  8  8  2  8  8 cos     ..... 0.906 (with no errors seen) can have M1A1 but errors rearranging result in M1A0 2

2

2

(b) M1: Uses formula for arc length with r = 8 and any angle i.e. s = 8 if working in rads or s =

 360

 2  8 in degrees

(If the formula is quoted with r the 8 may be implied by the value of their r  ) M1: Uses angles on straight line (or other geometry) to find angle BOC or AOD and uses Perimeter = 23 + arc lengths BC and AD (may make a slip – in calculation or miscopying) A1: correct work leading to awrt 40.9 not 40.8 (do not need to see cm) This answer implies M1M1A1 (c) Way 1: M1: Mark is given for correct statement of area of triangle 12  8  8  sin 0.906 (must use correct angle) or for co...